question evaluate the integral below. ∫(4ln(4x^4)/x^4)dx be sure to place the argument of any trigonometric…

Explanation:

Step1: Simplify the integrand

First, use the property of logarithms $\ln(ab)=\ln a+\ln b$. So, $\ln(4x^{4})=\ln 4 + 4\ln x$. Then the integrand $\frac{4\ln(4x^{4})}{x^{4}}=\frac{4(\ln 4 + 4\ln x)}{x^{4}}=\frac{4\ln 4}{x^{4}}+\frac{16\ln x}{x^{4}}$.

Step2: Split the integral

$\int\frac{4\ln(4x^{4})}{x^{4}}dx=\int\frac{4\ln 4}{x^{4}}dx+\int\frac{16\ln x}{x^{4}}dx$. The first - integral: $\int\frac{4\ln 4}{x^{4}}dx = 4\ln 4\int x^{-4}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $4\ln 4\int x^{-4}dx=4\ln 4\frac{x^{-4 + 1}}{-4 + 1}=-\frac{4\ln 4}{3x^{3}}$.

Step3: Solve the second integral using integration by parts

Let $u = \ln x$ and $dv=x^{-4}dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{x^{-4 + 1}}{-4+1}=-\frac{1}{3x^{3}}$. By the integration - by - parts formula $\int u;dv=uv-\int v;du$, we have $\int\frac{16\ln x}{x^{4}}dx=16\left(-\frac{\ln x}{3x^{3}}-\int\left(-\frac{1}{3x^{3}}\right)\frac{1}{x}dx\right)=16\left(-\frac{\ln x}{3x^{3}}+\frac{1}{3}\int x^{-4}dx\right)=16\left(-\frac{\ln x}{3x^{3}}+\frac{1}{3}\cdot\frac{x^{-3}}{-3}\right)=-\frac{16\ln x}{3x^{3}}-\frac{16}{9x^{3}}$.

Step4: Combine the results

$\int\frac{4\ln(4x^{4})}{x^{4}}dx=-\frac{4\ln 4}{3x^{3}}-\frac{16\ln x}{3x^{3}}-\frac{16}{9x^{3}}+C=-\frac{4\ln 4 + 16\ln x}{3x^{3}}-\frac{16}{9x^{3}}+C$.

Answer:

$-\frac{4\ln 4+16\ln x}{3x^{3}}-\frac{16}{9x^{3}}+C$